Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $15.3$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $25.2$ years.
Solution: $15.3$ $12$ $18.6$ $8.7$ $21.9$ $5.4$ $25.2$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $15.3$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $12$ years and one standard deviation above the mean is $18.6$ years. Two standard deviations below the mean is $8.7$ years and two standard deviations above the mean is $21.9$ years. Three standard deviations below the mean is $5.4$ years and three standard deviations above the mean is $25.2$ years. We are interested in the probability of a meerkat living less than $25.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $5.4$ years and the other half $({0.15\%})$ will live longer than $25.2$ years. The probability of a particular meerkat living less than $25.2$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.